Aspire Faculty ID #17945 · Topic: JEE Main 2026 (22 January Morning Shift) · Just now
JEE Main 2026 (22 January Morning Shift)

If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ is $Q(5, b, c)$, then $a^2 + b^2 + c^2$ is equal to

Solution


Point $M = \left(3,; \frac{b+1}{2},; \frac{c+a}{2}\right)$ satisfies the line

$\frac{3-6}{3} = \frac{\frac{b+1}{2}-7}{2} = \frac{\frac{c+a}{2}-7}{-2}$

$\Rightarrow -1 = \frac{b-12}{4} = \frac{c+a-14}{-4}$

$\Rightarrow b = 8 \quad ...(1)$

$c + a = 18 \quad ...(2)$

Now $PQ \perp L$

$\Rightarrow (4\hat{i} + (b-2)\hat{j} + (c-a)\hat{k}) \cdot (3\hat{i} + 2\hat{j} - 2\hat{k}) = 0$

$\Rightarrow 12 + 2(b-2) - 2(c-a) = 0$

$\Rightarrow 6 + (b-2) - (c-a) = 0$

$\Rightarrow b - c + a + 4 = 0$

$\Rightarrow 8 - c + a + 4 = 0$

$\Rightarrow c - a = 12 \quad ...(3)$

From (2) & (3):

$a = 3,; c = 15$

$\Rightarrow a^2 + b^2 + c^2 = 9 + 64 + 225 = 298$

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